Rational Man and Irrational Man both buy new cars, and they decide to drive around two racetracks from time $t = 0$ to $t = \infty.$  Rational Man drives along the path parameterized by
\begin{align*}
x &= \cos t, \\
y &= \sin t,
\end{align*}and Irrational Man drives along the path parameterized by
\begin{align*}
x &= 1 + 4 \cos \frac{t}{\sqrt{2}}, \\
y &= 2 \sin \frac{t}{\sqrt{2}}.
\end{align*}If $A$ is a point on Rational Man's racetrack, and $B$ is a point on Irrational Man's racetrack, then find the smallest possible distance $AB.$
Explanation: Rational Man's racetrack is parameterized by $x = \cos t$ and $y = \sin t.$  We can eliminate $t$ by writing
\[x^2 + y^2 = \cos^2 t + \sin^2 t = 1.\]Thus, Rational Man's racetrack is the circle centered at $(0,0)$ with radius 1.

Irrational Man's racetrack is parameterized by $x = 1 + 4 \cos \frac{t}{\sqrt{2}}$ and $y = 2 \sin \frac{t}{\sqrt{2}}.$  Similarly,
\[\frac{(x - 1)^2}{16} + \frac{y^2}{4} = \cos^2 \frac{t}{\sqrt{2}} + \sin^2 \frac{t}{\sqrt{2}} = 1.\]Thus, Irrational Man's race track is the ellipse centered at $(1,0)$ with semi-major axis 4 and semi-minor axis 2.

Let $O = (0,0),$ the center of the circle.

[asy]
unitsize(1 cm);

pair A, B, O;

path rm = Circle((0,0),1);
path im = shift((1,0))*yscale(2)*xscale(4)*rm;

O = (0,0);
A = dir(120);
B = (1 + 4*Cos(100), 2*Sin(100));

draw(rm,red);
draw(im,blue);
draw(A--B--O--cycle);

dot("$A$", A, NW);
dot("$B$", B, N);
dot("$O$", O, S);
[/asy]

By the Triangle Inequality, $OA + AB \ge OB,$ so
\[AB \ge OB - OA = OB - 1.\]If $B = (x,y),$ then
\[\frac{(x - 1)^2}{16} + \frac{y^2}{4} = 1,\]so $y^2 = -\frac{x^2}{4} + \frac{x}{2} + \frac{15}{4}.$  Then
\[OB^2 = x^2 + y^2 = \frac{3x^2}{4} + \frac{x}{2} + \frac{15}{4} = \frac{3}{4} \left( x + \frac{1}{3} \right)^2 + \frac{11}{3}.\]This is minimized when $x = -\frac{1}{3},$ in which case $OB = \sqrt{\frac{11}{3}} = \frac{\sqrt{33}}{3}.$

If we take $A$ as the intersection of $\overline{OB}$ with the circle, then
\[AB = OB - 1 = \boxed{\frac{\sqrt{33} - 3}{3}}.\]